Question

The van't Hoff factor for weak electrolyte $A_x{B}_y$, if $(n=x+y)$ can be given as:

• A. $i=1-\alpha +x\alpha +y\alpha$
• B. $i=1+(x+y)$ at infinite dilution
• C. $i=(n-1)\frac{A_v}{A_{\infty }}+1$
• D. Both A and C

Answer 1

The correct option is D Both A and C

$A_x{B}_y\leftrightharpoons A_x+B_y$

$i=1-\alpha +x\alpha +y\alpha$

$\therefore \alpha =\frac{i-1}{(x+y)-1}=\frac{i-1}{n-1}$

$=\frac{A_v}{A_{\infty }}=\frac{i-1}{n-1}$

$\therefore i=(n-1)\frac{A_v}{A_{\infty }}+1$

Also at infinite dilution, the degree of dissociation approaches 1.

$i=1-\alpha +x\alpha +y\alpha =1-1+(x+y)$

$=0+(x+y)$

Thus, options A and C are correct.